Let p be a fixed positive integer. Prove that the mapping f : Z → Z given by f(n) = { n+p if n is divisible by p, n if n is not divisible by p, is a bijection, and determine f^−1.

Question

Let p be a fixed positive integer. Prove that the mapping [latex]f : \mathbb{Z} → \mathbb{Z}[/latex] given by

[latex]f(n)={\left\{\begin{array}{l l}{n+p}&{{\mathrm{if}}}&{n{\mathrm{~is~divisible~by~}}p,}\ {n}&{{\mathrm{if}}}&{n{\mathrm{~is~not~divisible~by~}}p,}\end{array} ight.}[/latex]

is a bijection, and determine [latex]f^{-1}.[/latex]

Accepted Answer

Note from the definition of f that f(n) is divisible by p if and only if n is divisible by p. Thus if [latex]f(n_{1})=f(n_{2})[/latex] then either [latex]f(n_{1})[/latex] is divisible by p, in which case [latex]n_{1}=f(n_{1})-p=f(n_{2})-p=n_{2},\ \mathrm{or}\ \ f(n_{1})[/latex] is not divisible by p, in which case [latex]n_{1}=f(n_{1})=f(n_{2})=n_{2}.[/latex] Hence f is an injection.

To show that f is also surjective, suppose that [latex]k\in\mathbb{Z}.[/latex] If k is not divisible by p then f(k) = k; and if k is divisible by p then so is k - p whence f(k- p) = (k - p) + p = k. Hence f i s a surjection.

As for [latex]f^{-1}\colon\mathbb{Z} o\mathbb{Z},[/latex] this is given by

[latex]f^{-1}(n)={\left\{\begin{array}{l l}{n}&{{\mathrm{if~}}n{\mathrm{~is~not~divisible~by~}}p;}\ {n-p}&{{\mathrm{if~}}n{\mathrm{~is~divisible~by~}}p.}\end{array} ight.}[/latex]

Question 3.34: Let p be a fixed positive integer. Prove that the mapping f ......

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